You may also like

In this article, we look at solids constructed using symmetries of
their faces.

Euler's Formula and Topology

Stage: 5

Article by Alan Beardon

Published December 2000,February 2011.

Euler's formula for the sphere

Roughly speaking, a network (or, as mathematicians
would say, a graph ) is a collection of points, called
vertices , and lines joining them, called edges .
Each edge meets only two vertices (one at each of its ends), and
two edges must not intersect except at a vertex (which will then be
a common endpoint of the two edges). The collection of edges form
the boundary of certain areas, called faces . Faces must
not have holes in them or handles on them. If two faces have common
boundary points, then they must share a common edge (and only
this), or a common vertex (and only this).

In Figure 1 there are two examples of networks, and these
have
3 faces, 12 edges and 10 vertices,
and
(ii) 2 faces, 6 edges and 5 vertices,
respectively.

Figure 1

It is not necessary to assume that the lines in a network are
`straight lines'; they can be `curves' of any sort except that they
must not cross over themselves or each other. Moreover, we can draw
such networks on any surface (for example, the plane, a sphere, a
cylinder, and so on).

A triangulation of a surface is a network on the
surface all of whose faces are triangular (that is, they are
bounded by three edges). In fact, surveyors map the countryside by
triangulating it with `trig points' (usually on mountain tops) and
measuring angles and distances between these trig points. In this
case the trig points are the vertices of the triangulation.

The simplest example of a triangulation on the sphere (which we
think of as the surface of the earth) is found by drawing the
equator and, say, $n$ lines of longitude. In this example, which is
illustrated in Figure 2, there are $2n$ triangular faces, $n+2$
vertices ($n$ on the equator, and one at each pole), and $3n$
edges, so that $$\hbox{(number of faces)}-\hbox{(number of edges)}+
\hbox{(number of vertices)} = 2n - 3n + (n + 2) = 2.$$

Figure 2

It is perhaps surprising that this answer does not depend on
the choice of $n$ . Even more remarkable is the fact, known as
Euler's Theorem , that this formula holds for
ALL triangulations of the sphere. Leonard Euler
(1707-1783) was a Swiss mathematician who was, perhaps, the most
productive mathematician of all time. Thus, for any triangulation
of the sphere with, say, $T$ triangles, $E$ edges and $V$ vertices,
Euler's formula for the sphere is that $$T-E+V = 2.$$

The important thing to realise is that this formula is a
topological invariant : this means that if we deform the
triangulation and the sphere continuously then the numbers $T$, $E$
and $V$ will not change and the formula will still be true. For
example, as we can deform the sphere to a cube, the formula will
still hold for a cube, so let us check this with one particular
triangulation of the cube. We divide each face of the cube into two
triangles by drawing a diagonal across each face. Then there are
six faces of the cube and each gives two triangles; thus $T=12$.
Clearly $V=8$ (there are no new vertices introduced when we draw
the diagonals), and $E = 18$ (twelve from the original cube and the
six added diagonals). Thus $$T-E+V = 12-18+8 = 2.$$

EXERCISE 1

Check that Euler's formula is true in each of the following cases
(and in each case you should try different triangulations):

for a tetrahedron (a pyramid with a triangular base);

for the surface formed by gluing two tetrahedra together across
their base;

for a pyramid with a square base as found in Egypt (remember
that the base of the pyramid, where the pyramid touches the ground,
is part of the surface of the pyramid);

for the octahedron (which is formed by gluing two pyramids
together across their base);

the surface formed by gluing two pyramids, each with a square
base, onto two different faces of a cube. You should assume that
the faces of the cube are the same size as the bases of the
pyramids.

EXERCISE 2

Construct your own surface (for example, by sticking several cubes
together), triangulate it, and then check that Euler's formula
works for this.

Remark 1:

`Construct' here does not mean that you should physically construct
the surface; part of mathematics is learning how to represent three
dimensional objects on a sheet of paper!

Remark:

the surface must be a deformed sphere; for example, Euler's formula
as given above will not hold on the surface of a doughnut (a
torus).

Legendre's proof of Euler's formula on a sphere

The geometry of the sphere is extremely important; for example,
when navigators (in ships or planes) work out their course across
one of the oceans they must use the geometry of the sphere (and not
the geometry of the plane!). In the usual plane Euclidean geometry,
the angle sum of a triangle is $180$ degrees (or $\pi$ radians). On
the surface of the sphere, the arc of shortest distance between two
points is the arc of a great circle on the sphere, and if we form a
`spherical' triangle with these arcs as sides then the angle sum of
the triangle is NOT $180$ degrees. For example, if we look at the
triangle on the sphere bounded by the equator, the Greenwich
meridian, and the line of longitude given by $90^\circ$ East, we
obtain a spherical triangle all three of whose angles are
$90^\circ$. Let us call this triangle $\Delta$. You may like to
consider now what other angle sums of a triangle are possible on
the sphere.

It is a fact, which we shall assume here, that if a spherical
triangle is lying on a sphere of unit radius and has, say, angles
$\theta_1$, $\theta_2$ and $\theta_3$ (measured in radians), then
the area of the triangle is $\theta_1+\theta_2+\theta_3 -
\pi$. Of course, in Euclidean geometry, we would have
$\theta_1+\theta_2+\theta_3 = \pi$, but this formula is no longer
valid because our space is curved. At its simplest level,
Einstein's theory about space-time is that the space-time we live
in is curved and so the `formulae' of physics are different to
those that we get if we just use the Euclidean point of view.

Notice that if we accept the above formula for the area of a
spherical triangle, then the triangle $\Delta$ constructed above
has area $3(\pi/2)-\pi = \pi/2$. As eight copies of $\Delta$ will
fill the sphere without overlapping, we see that the area of
the sphere (of unit radius) is $4\pi$.

We can now give Legendre's beautiful proof of Euler's formula
that is based on a simple discussion of geometry on the sphere. For
any triangle on the sphere, we have $$\hbox{angle sum of triangle }
= \hbox{area of triangle} + \pi.$$ Suppose that there are $T$
triangles, $E$ edges and $V$ vertices in the triangulation of the
sphere. Then, summing all angles in all triangles, the total angle
sum is $2\pi V$ (as all of the angles occur at a vertex without
overlap, and the angle sum at any one of the $V$ vertices is
exactly $2\pi$). Also, the sum of the areas of the triangles is the
area of the sphere, namely $4\pi$; thus we see that $$2\pi V = 4\pi
+ T\pi ,$$ or $$V - T/2 = 2.$$ Now (by counting edges of each
triangle and noting that this counts each edge twice), we obtain
$3T=2E$, or $E=3T/2$. Thus we have $$T - E + V = T - 3T/2 + (2 +
T/2)= 2$$ which is Euler's formula.

Euler's formula for a simple closed polygon

Given a polygon that does not cross itself, we can triangulate
the inside of the polygon into non-overlapping triangles such that
any two triangles meet (if at all) either along a common edge, or
at a common vertex. Suppose that there are $T$ triangles, $E$ edges
and $V$ vertices; then Euler's formula for a polygon is
$$T-E+V = 1.$$ You should be able to confirm this in the polygon
illustrated in Figure 3.

Figure 3

EXERCISE 3

Draw different polygons, triangulate them, and check that Euler's
formula is true.

Proof of Euler's formula for a simple closed polygon

.

Because the terms $T$, $E$ and $V$ are unchanged when we apply a
continuous change (or deformation) to the picture, we can imagine
the picture to be drawn on a flexible rubber sheet. We cut the
polygon out of the sheet, and then manipulate it until it fits
exactly onto the lower half of a sphere sitting on the table. We
can now regard it as a triangulation of the lower hemisphere of
this sphere, and we shall use geographical terms, like `equator'
and `north pole' to describe the points on the sphere. There will
be a certain number of vertices, say $N$, on the `equator' of the
sphere (exactly the same number as were on the boundary of the
original polygon) and, as the vertices and edges alternate around
the equator, there will also be exactly $N$ edges on the equator.
Now draw arcs of circles from the `North pole' of the sphere to
each of these $N$ vertices (see Figure 4). This will now give us a
triangulation of the sphere with $N$ new triangular faces, $N$ new
edges (all from the North pole) and one new vertex (at the North
pole). Thus, from Euler's formula for the sphere, $$(T+N) - (E+N) +
(V+1) = 2,$$ and this gives Euler's formula for the polygon, namely
$$T-E+V = 1.$$

Figure 4

Euler's formula with polygonal faces on any surface

So far we have considered Euler's formula on a surface with the
network only having triangular faces. In fact, the formula also
holds when the faces are polygons. You should try some examples of
this to persuade yourself that this is so. As one example, the cube
has six square faces, 12 edges and 8 vertices and 6 - 12 + 8 =
2.

Here is a sketch of a proof of Euler's formula for a sphere $$F
- E + V = 2.$$ Some details are omitted but the general idea should
be clear.

We start with a polygon drawn on the sphere. Suppose this has
$N$ vertices and hence $N$ sides on the boundary of the polygon.
There are two faces (the inside and outside of the polygon) so for
this network we have $$F - E + V = 2 - N + N = 2.$$ Suppose we now
join two vertices by a polygonal curve - shown as a red line in the
diagram.

For a general curve of this type we add $k$ new edges, $k - 1$
new vertices (in the figure, $k = 4$), and the number of faces is
increased by one (one face is divided into two). Then for the new
network $$F_{new} - E_{new} + V_{new} = (F + 1) - (E + k) + (V + k
- 1)= F - E + V = 2.$$ In the same way we can add another 'dotted'
line without changing $F - E + V $, and in this way we can build a
general network for which we must still have $F - E + V = 2.$

For a polygon the proof is very much like the proof in Section
3.

A Euler's formula for a polygon with holes

Consider two polygons, one inside the other; some examples are
given in Figure 5. What is Euler's formula for the region
between the two polygons ? You should draw different
regions of this type, triangulate them in different ways, and when
you have reached an answer experimentally, try and prove it.

Figure 5

Now try to find Euler's formula for regions with two or more
holes like those in Figure 6.

Figure 6

A Euler's formula for a square tube

Consider a square tube with no ends as illustrated in Figure 7,
and consider triangulations of this that cover the whole tube.
What is Euler's formula for a square tube ? You
should draw different triangulations of this, and when you have
reached an answer experimentally, try and prove it.

Figure 7

When you have done this, try and find Euler's formula
for a `doughnut' : see Figure 8.

Figure 8

The angle deficiency of a polyhedron

Here is an attractive application of Euler's Formula. The angle
deficiency of a vertex of a polyhedron is $360^\circ $ (or $ 2\pi $
radians) minus the sum of the angles at the vertex of the faces
that meet at the vertex. For example, for a cube, there are three
faces, each with a right angle at each vertex so the angle
deficiency (at each vertex) is $ 2\pi - 3\times (\pi/2)\ = \pi /2 $
($ 360^\circ \ - 3\times90^\circ = 90^\circ $), or 90 degrees. The
total angle deficiency for the cube is $ 8\times 90^\circ\ =
720^\circ $ (or $4\pi$ radians). It is a remarkable fact that
the total angle deficiency of any polyhedron that is
topologically equivalent to a sphere is $720^\circ$ or $4\pi$
radians .

Suppose that we have a polyhedron which is made up with faces
$F_1,\ldots F_k$ (each of these is a polygon, which need not be
regular). Then we can apply Euler's Theorem to the polyhedron, so
let us count the faces, edges and vertices. First, by definition,
there are $k$ faces. Suppose that the face $F_j$ has $N_j$ edges
(and hence $N_j$ vertices). If we count the total number of edges
by looking at them from inside each face we `see' $N_1+\cdots +N_k$
edges. But of course, we see each edge twice (for each edge is
`seen' from both sides); thus $$2E = N_1+\cdots + N_k.$$ One
interesting fact here is that $N_1+\cdots +N_k$ must be
even ; this is not obviously so. Finally, the number $V$ of
vertices is given by Euler's formula for a sphere (or for any
`deformed sphere'), so $$V = E-F+2 = (N_1+\cdots + N_k)/2 - k +2,$$
or $$2V = (N_1+\cdots + N_k) - 2k +4.$$

The polyhedron has $V$ vertices and, by definition, the
total angle deficiency of the polyhedron is $2\pi V$ minus the
sum of the angles of all faces at all vertices . As the
interior angle sum of a polygon with $n$ sides is $(n-2)\pi$, we
see that the total angle deficiency of the polyhedron is $\Theta$,
where

Theorem

.

The total angle deficiency of any polyhedron is $4\pi$.

An example

Consider the regular tetrahedron. This has three equilateral
triangles meeting at each of four vertices, so in this case,
$$\Theta = 8\pi - 4\times (3\times \pi/3) = 4\pi.$$ You may like to
try this with other examples.

The fact that the total angle deficiency of a polyhedron is 720
degrees, together with Euler's formula, gives the key to finding
how many regular polyhedra there are (Platonic Solids) and how many
semi-regular polyhedra there are (Archimedean solids) and
discovering their properties (the shapes and number of faces
etc.).

Adding Euler numbers

We are going to show that if we join two surfaces together
across the boundary of a hole in each, then the Euler number of the
joined surface is the sum of the Euler numbers of the two separate
surfaces .

Suppose that we have two surfaces $S_1$ and $S_2$, each with a
hole in it. By deforming the surfaces, we may assume that the two
holes are circular with the same radius; for example as illustrated
below.

Now triangulate both surfaces in such a way that there are
exactly $k$ vertices, and hence $k$ edges, on each of the circular
boundaries of the holes. Again by deforming the surfaces we may
assume that these $k$ vertices and $k$ edges match up exactly when
the two surfaces are brought together.

Suppose that the triangulation of $S_1$ has $T_1$ triangles,
$E_1$ edges and $V_1$ vertices, and similarly for $S_2$. When the
surfaces have been joined together at the edges of the two circular
holes, we will have a triangulation of the new surface, which we
call $S$, and this triangulation will have exactly $T_1+T_2$
triangles in it. However, it will not have $E_1+E_2$ edges because
each of the $k$ edges on the boundary of one hole will be joined to
one of the other edges on the other boundary; thus the new
triangulation will have exactly $E_1+E_2-k$ edges and, similarly,
$V_1+V_2-k$ vertices. Thus $$ \begin{eqnarray}
\hbox{Euler-number}(S) &=(T_1+T_2) - (E_1+E_2-k) +
(V_1+V_2-k)\\ &=(T_1-E_1+V_1) + (T_2-E_2+V_2)\\ &=
\hbox{Euler-number}(S_1) + \hbox{Euler-number}(S_2). \end{eqnarray}
$$

As an example, a disc is topologically a hemisphere, so that
these two surfaces have the same Euler number. If we join two
hemispheres across their boundaries (for example, the southern and
northern hemishperes are joined across the equator) we see that the
Euler number of a sphere is twice the Euler number of a hemisphere.
Hence the Euler number of a hemisphere, and also of a disc, is
one.

The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the
NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to
embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities
can be found here.